Answer: 0.049 mol
Explanation:
1) Data:
n₁ = 0.250 mol
p₁ = 730 mmHg
p₂ = 1.15 atm
n₂ - n₁ = ?
2) Assumptions:
i) ideal gas equation: pV = nRT
ii) V and T constants.
3) Solution:
i) Since the temperature and the volume must be assumed constant, you can simplify the ideal gas equation into:
pV = nRT ⇒ p/n = RT/V ⇒ p/n = constant.
ii) Then p₁ / n₁ = p₂ / n₂
⇒ n₂ = p₂ n₁ / p₁
iii) n₂ = 1.15atm × 760 mmHg/atm × 0.250 mol / 730mmHg = 0.299 mol
iv) n₂ - n₁ = 0.299 mol - 0.250 mol = 0.049 mol
I would say its c
given a fixed volume, when volume decreases while mass remains constant
density increases ie temperature increases also
Explanation:
The work done equals the change in energy.
W = ΔKE
W = 0 − ½mv²
W = -½ (0.270 kg) (-7.50 m/s)²
W = -7.59 J
Work is force times displacement.
W = Fd
-7.59 J = F (-0.150 m)
F = 50.6 N
Answer: CaF2
Explanation:
Calcium is a metal and has 2+ cation charge. While F us in group 7 with an oxidation of -1.
So. Ca²+ F- do the criss cross rule where the charge of the cation will be the subscript of the anion and vice versa. So the result is CaF2
Answer:
Approximately 
assuming no heat exchange between the mixture and the surroundings.
Explanation:
Consider an object of specific heat capacity 
and mass 
. Increasing the temperature of this object by 
would require 
.
Look up the specific heat of water: 
.
It is given that the mass of the water in this mixture is 
.
Temperature change of the water: 
.
Thus, the water in this mixture would have absorbed :
.
Thus, the energy that water absorbed was: 
.
Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, 
, would be the opposite of the energy that the metal in this mixture released.
Thus: 
(negative because the metal in this mixture released energy rather than absorbing energy.)
Mass of the metal in this mixture: 
.
Temperature change of the metal in this mixture: 
.
Rearrange the equation to obtain an expression for the specific heat capacity: 
. The (average) specific heat capacity of the metal pieces in this mixture would be:
.
